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STRUCTURE OF CHLORINE ISOTOPES
.By Prof. Lefteris Kaliambos (Natural Philosopher in New Energy) (August 2014) Historically the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories, which could not lead to the nuclear structure. Under this physics crisis and using the charged UP and DOWN quarks discovered by Gell-Mann and Zweig I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” (2003), which led to my discovery of the new structure of protons and neutrons given by proton = + 5d + 4u = 288 quarks = mass of 1836.15 electrons neutron = + 4u + 8d = 288 quarks = mass of 1838,68 electrons The paper was also presented at a nuclear conference held at NCSR "Demokritos" (2002). Here one can see the 9 charged quarks in proton and the 12 ones in neutron able to give the charge distributions in nucleons for revealing the strong electromagnetic force for the nuclear binding in the correct nuclear structure by applying the laws of electromagnetism. You can see my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS . Note that according to my discovery of the LAW OF ENERGY AND MASS the mass defect in the nuclear structure is due to the photon mass of the emitting dipolic photon presented at the international conference "Frontiers of fundamental physics" (1993) organised by the natural philosophers M. Barone and F. Selleri , who gave me an award including a disc of the atomic philosopher Democritus. Nevertheless today many physicist continue to apply not the well-established laws but the various fallacious nuclear structure models which lead to complications. Chlorine (Cl) has 24 isotopes with mass numbers ranging from Cl-28 to Cl-51. There are two principal stable isotopes, Cl-35 (75.78%) and Cl-37 (24.22%), found in the relative proportions of 37.89:12.11, respectively, giving chlorine a standard atomic mass of 35.453. The longest-lived radioactive isotope is Cl-36 which has a half-life of 301,000 years. All other isotopes have half-lives under 1 hour, many less than one second. The shortest-lived are Cl-29 and Cl-30, with half-lives less than 20 and 30 nanoseconds, respectively—the half-life of Cl-28 is unknown. WHY Cl-35 AND Cl-37 WITH S = +3/2 ARE STABLE NUCLIDES After a careful analysis of the structure of atomic nuclei I discovered that the beta decay is due to the fact that in unstable nuclei there exist single horizontal pn bonds of weak binding energy leading to the beta decay. For example in my paper STRUCTURE AND BINDING OF H3 AND He3 using the diagram of the structure of the H3 one sees that it is unstable because the two neutrons make single np bonds, while the He3 is stable because the one neutron between the two protons makes two np bonds per neutron. On the other hand the pp repulsions of long range lead to the instability when we have a small number of pn bonds per nucleon. '' ''DIAGRAM OF Cl-35 WITH S =+3/2 ' . p12.......n12' ' -HP6 n11.......p11' ' . p17..... n10.......p10.....n18' ' +HP5 n17.......p9.......n9 ' ' . n14........p8........n8.......p16 ' ' -HP4 p14.......n7.......p7.......n16 ' ' . p13.......n6........p6.......n15' ' +HP3 n13.......p5.........n5.......p15 ' ' . p4.......n4' ' -HP2 n3.........p3 ' ' . n2........p2' ' +HP1 p1........n1 ' For comparing the structure of Cl-35 with Ar-35 you can read my STRUCTURE OF Cl-35 AND Ar-35 . In the following diagram of Cl-35 you can see that the deuterons like the p13 n13, p14n14 and p17n17 with S=+1 make strong vertical rectangles on the left side of the core (parallelepiped ofMag-24 with S=0) while the two deuterons like the p15n15 and p16n16 with S = 0 form a vertical rectangle on the right side of the core which becomes a stable one because the extra n18(+1/2) at the fifth horizontal plane of positive spins (+HP5) makes two bonds per neutron. Under such arrangements of nucleons the Cl-35 is a stable nuclide with S=+3/2 given by S = 0 +1 +0 +1/2 = +3/2 Note that all isotopes with S = +3/2 are based on the structure of Cl-35. For example in the stable structure of Cl-37 with S=+3/2 the deuteron p17n17 with S=+1 goes from the +HP5 to the +HP3 with the same S=+1 to make horizontal bonds with p6 and n6. They are not shown in a diagram because they exist in front of p6 and n6. In this arrangement since the p17 of the deuteron n17p17 and the p15 are at the same +HP3 they form a blank position able to receive the n19(+1/2). Also the p17 with the p2 of the -HP2 form a blank position able to receive the n20(-1/2) existing at the -HP2. That is, we observe a stable structure because the n18(+1/2) of the +HP5, the n19(+1/2) of the +HP3 and the n20(-1/2) of the -HP2 make two bonds per neutron. Thus, the spin S =+3/2 of the Cl-17 is given by S = +1 +1/2 +1/2 -1/2 =+3/2. In other words the stable Cl-37 has the same spin S=+3/2 as that of the stable Cl-35 because the two extra neutrons like the n19 and n20 have opposite spins. ' ' NUCLEAR STRUCTURE OF Cl-39, Cl-43, Cl-45, Cl-47, Cl-49 AND Cl-51 WITH S =+3/2 ' The structure of the above unstable isotopes is based on the structure of Cl-35 because the extra neutrons of opposite spins make single horizontal bonds leading to the beta decay. For example the Cl-51 with S =+3/2 has 16 extra neutrons of opposite spins which form single bonds and lead to the beta decay. ' ''' '''NUCLEAR STRUCTURE OF Cl-33, Cl-31 AND Cl-29 WITH S =+3/2 In the absence of neutrons with opposite spins we get also the structure of the above nuclides based on the structure of Cl-35 with S =+3/2. For example in the structure of Cl-33 with S=+3/2 we have two absent neutrons with opposite spins. Note that in the absence of neutrons the smaller number of bonds cannot overcome the pp repulsions of long range and lead to the decay. ' ' NUCLEAR STRUCTURE OF Cl-34, Cl-36, Cl-38, Cl-40, Cl-46, Cl-48 and Cl-50 In the following diagram of the unstable Cl-34 with S=+3 you see that the core is not the parallelepiped of Mg-24. Here the parallelepiped has 5 horizontal planes like the +HP1, -HP2, +HP3, -HP4 and +HP5 with S = +3, in which an unstable rectangle of the deuterons p16n16 and p17n17 with S=0 makes 4 weak horizontal bonds with the nucleons of the core. After a careful analysis I found that the structure of the above unstable nuclides with A>34 is based on the structure of Cl-34 in which the extra neutrons make single bonds leading to the beta decay . For example in the structure of Cl-36 with S =+2 the two extra neutrons have negative spins with a total S=-1. That is S = +3 -1 = +2 . However in the case of Cl-38 with S=-2 the nucleons of the structure of Cl-34 change the spins. For example we have the five horizontal planes as -HP1, +HP2, -HP3 +HP4 and -HP5 with S = - 3 . Thus, in Cl-38 with S =-2 we have two extra neutrons of negative spins and two neutrons of opposite spins. ' DIAGRAM OF Cl-34 WITH S = +3' ' . p15.....n10......p10' ' +HP5 n15.......p9.......n9 ' ' . n14.......p8........n8 ' ' -HP4 p14......n7........ p7 ' ' . p13.......n6........p6.......n17' ' +HP3 n13.......p5........n5.......p17 ' ' . n12........p4........n4........p16' ' -HP2 p12......n3........p3.......n16 ' ' . p11......n2........p2' ' +HP1 n11.....p1.......n1 ' ' ' NUCLEAR STRUCTURE OF Cl-30 AND Cl-28 In the absence of neutrons we get also the structure of the above unstable nuclides based on the structure of Cl-34 with S=+3 . For example in the structure od Cl-28 with S=+1 we have 4 absent neutrons with positive spins and 2 absent neutrons with opposite spins giving S=0. That is, the total spin is given by S = +3 -4(+1/2) +0 = +1. Category:Fundamental physics concepts